## Circular Gate with Water on One Side and Air on the Other Side

**Situation**

The figure below shows a vertical circular gate in a 3-m diameter tunnel with water on one side and air on the other side.

- Find the horizontal reaction at the hinge.

A. 412 kN

B. 408 kN

C. 410 kN

D. 414 kN - How far from the invert of the tunnel is the hydrostatic force acting on the gate?

A. 1.45 m

B. 1.43 m

C. 1.47 m

D. 1.41 m - Where will the hinge support be located (measured from the invert) to hold the gate in position?

A. 1.42 m

B. 1.46 m

C. 1.44 m

D. 1.40 m

## Three Reservoirs Connected by Pipes at a Common Junction

**Situation**

Three reservoirs *A*, *B*, and *C* are connected respectively with pipes 1, 2, and 3 joining at a common junction *P*. Reservoir *A* is at elevation 80 m, reservoir *B* at elevation 70 m and reservoir *C* is at elevation 60 m. The properties of each pipe are as follows:

*L*= 5000 m,

*D*= 300 mm

Pipe 2:

*L*= 4000 m,

*D*= 250 mm

Pipe 3:

*L*= 3500 m

The flow from reservoir *A* to junction *P* is 0.045 m^{3}/s and *f* for all pipes is 0.018.

- Find the elevation of the energy grade line at
*P*in m.

A. 75.512

B. 73.805

C. 72.021

D. 74.173 - Determine the flow on pipe 2 in m
^{3}/s.

A. 0.025

B. 0.031

C. 0.029

D. 0.036 - Compute the diameter appropriate for pipe 3 in mm.

A. 175

B. 170

C. 178

D. 172

## Reversed Curve to Connect Three Traversed Lines

**Situation**

A reversed curve with diverging tangent is to be designed to connect to three traversed lines for the portion of the proposed highway. The lines *AB* is 185 m, *BC* is 122.40 m, and *CD* is 285 m. The azimuth are Due East, 242°, and 302° respectively. The following are the cost index and specification:

Number of Lanes = Two Lanes

Width of Pavement = 3.05 m per lane

Thickness of Pavement = 280 mm

Unit Cost = P1,800 per square meter

It is necessary that the *PRC* (Point of Reversed Curvature) must be one-fourth the distance *BC* from *B*.

- Find the radius of the first curve.

A. 123 m

B. 156 m

C. 182 m

D. 143 m - Find the length of road from
*A*to*D*. Use arc basis.

A. 552 m

B. 637 m

C. 574 m

D. 468 m - Find the cost of the concrete pavement from
*A*to*D*.

A. P2.81M

B. P5.54M

C. P3.42M

D. P4.89M

## Problem 04 - Symmetrical Parabolic Curve

**Problem**

A highway engineer must stake a symmetrical vertical curve where an entering grade of +0.80% meets an existing grade of -0.40% at station 10 + 100 which has an elevation of 140.36 m. If the maximum allowable change in grade per 20 m station is -0.20%, what is the length of the vertical curve?

- 150 m
- 130 m
- 120 m
- 140 m

## Influence Lines for Beams

A downward concentrated load of magnitude 1 unit moves across the simply supported beam *AB* from *A* to *B*. We wish to determine the following functions:

- reaction at
*A* - reaction at
*B* - shear at
*C*and - moment at
*C*

when the unit load is at a distance *x* from support *A*. Since the value of the above functions will vary according to the location of the unit load, the best way to represent these functions is by influence diagram.

## Influence Lines

Influence line is the graphical representation of the response function of the structure as the downward unit load moves across the structure. The ordinate of the influence line show the magnitude and character of the function.

The most common response functions of our interest are *support reaction*, *shear at a section*, *bending moment at a section*, and *force in truss member*.

With the aid of influence diagram, we can...

- determine the position of the load to cause maximum value to the function.
- calculate the maximum value of the function.

Value of the function for any type of load

$\displaystyle \text{Function} = \int_a^b y_i (y \, dx)$

## Example 02: Notched beam with concentrated load

**Problem**

A 150 mm by 300 mm wooden beam having a simple span of 6 meters carries a concentrated load *P* at its midspan. It is notched at the supports as shown in the figure. For this problem, all calculations are based on shear alone using the 2010 NSCP specification given below. Allowable shear stress of wood, *F _{v}* = 1.0 MPa.

- If
*P*= 30 kN, calculate the maximum allowable depth (millimeters) of notches at the supports.- 88
- 62
- 238
- 212

- If the depth of notches is 100 mm, what is the safe value of
*P*(kiloNewton) the beam can carry.- 26.67
- 17.78
- 8.89
- 13.33

- If
*P*= 25 kN and the depth of notches is 150 millimeters, what is the shear stress (MegaPascal) near the supports.- 0.83
- 6.67
- 1.67
- 3.33

**NSCP 2010 Section 616.4: Horizontal Shear in Notched Beams**

When rectangular-shaped girder, beams or joists are notched at points of support on the tension side, they shall meet the design requirements of that section in bending and in shear. The horizontal shear stress at such point shall be calculated by:

Where:

$d'$ = actual depth of beam at notch

## Example 01: Safe uniform load for beam notched at tension side of support

**Problem**

A 75 mm × 150 mm beam carries a uniform load *w _{o}* over the entire span of 1.2 m. Square notches 25 mm deep are provided at the bottom of the beam at the supports. Calculate the safe value of

*w*based on shear alone.

_{o}Allowable shear normal to grain = 1.85 MPa

## Notching on Beams

**NSCP 2010**

When rectangular-shaped girder, beams or joists are notched at points of support on the tension side, they shall meet the design requirements of that section in bending and in shear. The horizontal shear stress at such point shall be calculated by:

Where:

$d'$ = actual depth of beam at notch.